91Ó°ÊÓ

Mass of coffee filter$\mathrm{m}=1.8\mathrm{g}$ , height$\mathrm{h}=4\mathrm{m}$ , speed$\mathrm{v}=0.8\mathrm{m}/\mathrm{s}$ .

The potential energy is given by:

$\mathbf{U}\mathbf{=}\mathbf{mgh}$

Here denotes mass. Theis acceleration due to gravity that is given by$\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$,h stands for height.

The kinetic energy is given by:

$\mathbf{k}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$

Here,is mass of the object andis the velocity of the object.

The mass of coffee filter is m=1.8g.

Convert the unit of mass into kg.

$$\begin{gathered} \mathrm{m}=1.8\mathrm{g}\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{g}}\right) \\ =1.8×{10}^{-3}\mathrm{kg} \end{gathered}$$

Thus, out of all, the observers in situation may feel that Newton’s first law has been violated.

Substitute , $\mathrm{m}=1.8×{10}^{-3}\mathrm{kg},\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2\mathrm{and}\mathrm{h}=4\mathrm{m}$into the formula of potential energy.

role="math" localid="1657795625494" $$\begin{gathered} \mathrm{U}=\left(1.8×{10}^{-3}\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(4\mathrm{m}\right) \\ =70.56×{10}^{-3}\mathrm{J} \end{gathered}$$

Therefore, the potential energy is $70.56×{10}^{-3}\mathrm{J}$.

Substitute$\mathrm{m}=1.8×{10}^{-3}\mathrm{kg}\mathrm{and}\mathrm{v}=0.8\mathrm{m}/\mathrm{s}$ andinto the formula of kinetic energy.

$$\begin{gathered} \mathrm{k}=\frac{1}{2}\left(1.8×{10}^{-3}\mathrm{kg}\right){\left(0.8\mathrm{m}/\mathrm{s}\right)}^2 \\ =0.576×{10}^{-3}\mathrm{J} \end{gathered}$$

Therefore, the kinetic energy is $0.576×{10}^{-3}\mathrm{J}$.

The loss of energy is given by $$\begin{gathered} \mathbf{∆}\mathbf{E}\mathbf{=}\mathbf{PE}\mathbf{-}\mathbf{KE} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$.

Subtract the obtained kinetic energy from the obtained potential energy to obtain the amount of energy loss.

$$\begin{gathered} \mathbf{∆}\mathbf{E}\mathbf{=}\mathbf{PE}\mathbf{-}\mathbf{KE} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{70}\mathbf{.}\mathbf{56}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{J}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{576}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{J} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{69}\mathbf{.}\mathbf{984}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{J} \end{gathered}$$

The loss of energy is said to be transferred in the air molecules.

Therefore, loss of kinetic energy is$69.984×{10}^{-3}\mathrm{J}$ and energy is transferred in air molecules.